Apply the three equations of motion for a rigid body in planar motion.

Analyze problems involving translational motion.

PLANAR KINETIC EQUATIONS OF MOTION: TRANSLATION

APPLICATIONS

The boat and trailer undergo rectilinear motion. In order to find the reactions at the trailer wheels and the acceleration of the boat, we need to draw the FBD and kinetic diagram for the boat and trailer.

How many equations of motion do we need to solve this problem? What are they?

APPLICATIONS (continued)

As the tractor raises the load, the crate will undergo curvilinear translation if the forks do not rotate.

If the load is raised too quickly, will the crate slide to the left or right?

How fast can we raise the load before the crate will slide?

PLANAR KINETIC EQUATIONS OF MOTION

• We will limit our study of planar kinetics to rigid bodies that are symmetric with respect to a fixed reference plane.

• As discussed in Chapter 16, when a body is subjected to general plane motion, it undergoes a combination of translation and rotation.

• First, a coordinate system with its origin at an arbitrary point P is established.

The x-y axes should not rotate and can either be fixed or translate with constant velocity.

• If a body undergoes translational motion, the equation of motion is F = m aG . This can also be written in scalar form as

Fx = m(aG)x and Fy = m(aG)y

EQUATIONS OF TRANSLATIONAL MOTION (continued)

• In words: the sum of all the external forces acting on the body is equal to the body’s mass times the acceleration of it’s mass center.

When a rigid body undergoes only rectilinear translation, all the particles of the body have the same acceleration so aG = a and = 0 in which case the rotational equation of motion applied at point G reduces to MG = 0. .The equations of motion become:

Note that, if it makes the problem easier, the moment equation can be applied about another point instead of the mass center. For example, if point A is chosen, MA = (m aG ) d .

EQUATIONS OF MOTION: RECTILINEAR TRANSLATION

Fx = m(aG)x

Fy = m(aG)y

MG = 0

The sum of all external forces acting on the body ( F) = the body’s mass (m) times the acceleration of the body’s mass center (aG).

PROCEDURE FOR ANALYSIS

Problems involving kinetics of a rigid body in only rectilinear translation should be solved using the following procedure:

1. Establish an (x-y) inertial coordinate system and specify the sense and direction of acceleration of the mass center, aG.

2. Draw a FBD and kinetic diagram showing all external forces.

3. Identify the unknowns.

5. Remember, friction forces always act on the body opposing the motion of the body.

4. Apply the three equations of motion (one set or the other):

Fx = m(aG)x Fy = m(aG)y

MG = 0

PROBLEM SOLVING

Given: The handcart has a mass of 200 kg and center of mass at G. A force of P=50 N is applied to the handle. Neglect the mass of the wheels.

Find: The normal reactions at each of the two wheels at A and B.

Plan: Follow the procedure for analysis.

PROBLEM SOLVING (continued)

Solution: The cart will move along a rectilinear path. Draw FBD and kinetic diagram.

=

x

y

Applying the equations of motion:

Fx = m(aG)x

50 cos 60 = 200 aG aG = 0.125 m/s2

+

MG = 0

-(0.3)NA+(0.2)NB+0.3(50 cos 60) – 0.6(50 sin 60) = 0

− 0.3 NA+ 0.2 NB = 18.48 N m (2)

+↑ Fy = 0 NA + NB –1962 –50 sin 60 = 0

NA + NB = 2005 N (1)

PROBLEM SOLVING (continued)

=x

y

Using Eqs. (1) and (2), solve for the reactions, NA and NB NA = 765 N, NB = 1240 N

Applying the equations of motion:

Given:A 50 kg crate rests on a horizontal surface for which the kinetic friction coefficient k = 0.2.

Find: The acceleration of the crate if P = 600 N.

Plan: Follow the procedure for analysis.

Note that the load P can cause the crate either to slide or to tip over. Let’s assume that the crate slides. We will check this assumption later.

EXAMPLE

The coordinate system and FBD are as shown. The weight of (50)(9.81) N is applied at the center of mass and the normal force Nc acts at O. Point O is some distance x from the crate’s center line. The unknowns are Nc, x, and aG .

EXAMPLE (continued)

Applying the equations of motion:

Solution:

Fx = m(aG)x: 600 – 0.2 Nc = 50 aG

Fy = m(aG)y: Nc – 490.5 = 0

MG = 0: -600(0.3) + Nc(x) – 0.2 Nc(0.5) = 0

Nc = 490 N

x = 0.467 m

aG = 10.0 m/s2

Since x = 0.467 m < 0.5 m, the crate slides as originally assumed.

If x was greater than 0.5 m, the problem would have to be reworked with the assumption that tipping occurred.

EXAMPLE (continued)